A container of fixed volume contains 8.72 × 1023helium atoms, each of mass 6.646

Question

A container of fixed volume contains 8.72 × 1023helium atoms, each of mass 6.646 × 1027 kg.You put the container in a freezer and decrease its temperature from 25.0C to -115 C.

Part A

What is the change in the thermal energy of the gas?

Express your answer with the appropriate units.

SubmitMy AnswersGive Up

Part B

What is the change in the entropy of the gas?

A container of fixed volume contains 8.72 × 1023helium atoms, each of mass 6.646 × 1027 kg.You put the container in a freezer and decrease its temperature from 25.0C to -115 C.

Part A

What is the change in the thermal energy of the gas?

Express your answer with the appropriate units.

SubmitMy AnswersGive Up

Part B

What is the change in the entropy of the gas?

Explanation / Answer

here

KE = thermal energy = 3/2 KT

K is Boltzman constant = 1.38 e -23 Joule/K mol

T1 = 25 + 273 = 298 K

T2 -115+ 273 = 158 K

change of energy dE = (E2-E1_

dE = 3/2* 1.38 e -23 * (298-158)

dE = 2.898 *10^-21 JOules /mole

change in energy of gas = 2.898 *10^-21 * 8.72 * 10^23

dE = 2527/06 JOules (ANSWER)

-----------------------------------------

Entropy dS = 3/2 n K ln (T2/T1)

dS = 3/2 * 8.72*10^23 * 1.38 *10^-23 * ln(158/298)

dS = -11.45 J/K (ANSWER)

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.